Question

Question #15909

Answer 1

The hybridisation of `P` IN `PCl_5` is `sp^3d` and the geometry of the compound is Trigonal bi-pyramidal

Explanation:

Hybridisation`=` Number of `sigma` bonds `+` Number of lone pairs.

In the case of `PCl_5` , `P` has `5` `sigma` bonds and `0` lone pairs.

So, the hybridisation is `5`.

Therefore it is `sp^3d` hybridized.

Now compounds with `sp^3d` hybridisation have Trigonal bi-pyramidal geometry.

There are `3` `Cl` bonds in the equitorial plane at an angle of `120` degrees from each other and `2` `Cl` bonds in the axial plane at an angle of `90` degrees from the equitorial plane.